#include "mbed.h

DigitalOut LED1(p10);
DigitalOut LED2(p11);
DigitalIn Switch(p12);

int main() {
	
	while (1) {
		
		if (Switch == 0) {
			LED1 = 0;     // Turn led1 off.
			LED2 = 1;     // Turn led2 on
			wait(0.5);
			LED2 = 0;
			wait(0.5);
		}

		
		else if (Switch == 1) {
			LED2 = 0;    
			LED1 = 1;  
			wait(0.5);
			LED1 = 0;
			wait(0.5);
		}
	}
}

Hello, i am new to MBED's and i am trying to follow a simple tutorial but i am having trouble understanding and copying the demonstration as i do not want to run the program and destroy any hardware by placing the resistor in the incorrect place. Sorry if this is a dumb question but i am only 15 and i am interested in electronics but i do not want to destroy any of the hardware. Any help would be greatly appreciated. Here is my program below which i am trying to build, but i am having trouble with the hardware: <<code title=Program 1>> #include "mbed.h DigitalOut LED1(p10); DigitalOut LED2(p11); DigitalIn Switch(p12); int main() { while (1) { if (Switch == 0) { LED1 = 0; // Turn led1 off. LED2 = 1; // Turn led2 on wait(0.5); LED2 = 0; wait(0.5); } else if (Switch == 1) { LED2 = 0; LED1 = 1; wait(0.5); LED1 = 0; wait(0.5); } } } <</code>> I am trying to replicate this diagram here but to the ports i specified in my program: {{/media/uploads/idannnx/capture.png}} So for my program am i right in assuming: > The center of the switch is placed in p12 of the mbed. > Connect LED1's cathode to p10 and the anode to the current limit resistor - inline with 'G' on the breadboard > Connect LED2's cathode to p11 and the anode to the current limit resistor - inline with 'G' on the breadboard > Connect a wire directly from p12 to vout (Opposite side of breadboard (3.3v out)) > Connect a wire directly from p11 to gnd (0v) > Connect a wire directly from p10 to gnd (0v) > Current a resistor to the same pin as both of the anodes of p11 and p12, and the otherside of the resistor as shown in the image. (E6)

Hmmm...your assumptions are wrong. First, look at the pinout diagram carefully here: https://developer.mbed.org/platforms/mbed-LPC1768/ See where P12 is? It's not at the center of the switch. It's several pins below. Your code is driving the outputs P10, P11 high (3.3V) when you want to turn the LEDs on. Never connect those pins directly to ground or you can damage your part if you turn the output on. So...connect the ANODE of the LED to P10, then the cathode of the LED to one side of the resistor and the other side of the resistor to ground. Same goes for P11. It looks like you want to run an external switch to P12? An easy way to do this is to connect a resistor between P12 and 3.3V. The resistor needs to be something like 10,000 ohms. Then connect the switch to P12 on one side of the switch and ground on the other. Now when you move the switch it will connect P12 to "0" (ground) and "1" (3.3V via the resistor). Hope that helps.